Question

prove identity
where the angle involved acute angle
​

Answer 1

${\huge{\underline{\small{\mathbb{\purple{Ans~♥~}}}}}}$

cosA-sinA+1/cosA +sinA-1 = cosecA+cotA

Use the formula cot^2A-cosec^2A

LHS- cosA-sinA+1/cosA+sinA-1

divide by sinA on both numerator and denominator

cosA/sinA-sinA/sinA+1/sinA /cosA/sinA-sinA/sinA+1/sinA

cotA-1+cosecA/cotA+1-cosecA

at the place of 1 put above formula

cotA+cosec-(cot^2A-cosec^2A)/cotA-cosecA+1

now put it in a^2-b^2 =(a+b)(a-b)

cotA+cosecA+(cotA-cosecA)(cotA+cosecA)/cotA-cosecA+1

now common taking from numerator

cotA+cosecA(1+cotA-cosecA)/cotA-cosecA+1

Hence it prove cotA+cosecA

Answer 2

${\huge{\underline{\small{\mathbb{\pink{♡Ans:-}}}}}}$

cosA-sinA+1/cosA +sinA-1 = cosecA+cotA

Use the formula cot^2A-cosec^2A

LHS- cosA-sinA+1/cosA+sinA-1

divide by sinA on both numerator and denominator

cosA/sinA-sinA/sinA+1/sinA /cosA/sinA-sinA/sinA+1/sinA

cotA-1+cosecA/cotA+1-cosecA

at the place of 1 put above formula

cotA+cosec-(cot^2A-cosec^2A)/cotA-cosecA+1

now put it in a^2-b^2 =(a+b)(a-b)

cotA+cosecA+(cotA-cosecA)(cotA+cosecA)/cotA-cosecA+1

now common taking from numerator

cotA+cosecA(1+cotA-cosecA)/cotA-cosecA+1

Hence it prove cotA+cosecA